Question 1180207
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By basic rules for decomposing fractions, the decomposition is of the form<br>
{{{(x-2)/((x+2)^2(x-4)) = A/(x+2)+B/(x+2)^2+C/(x-4)}}}<br>
Multiply by the least common denominator:<br>
{{{x-2=A(x+2)(x-4)+B(x-4)+C(x+2)^2}}}<br>
{{{x-2=A(x^2-2x-8)+B(x-4)+C(x^2+4x+4)}}}<br>
{{{x-2=(A+C)x^2+(-2A+B+4C)x+(-8A-4B+4C)}}}<br>
Equate the coefficients on the two sides of the equation:<br>
(1) {{{A+C=0}}}
(2) {{{-2A+B+4C=1}}}
(3) {{{-8A-4B+4C=-2}}}<br>
Solve (1) to get C=-A and substitute in (2) and (3).<br>
(4) {{{-6A+B=1}}}
(5) {{{-12A-4B=-2}}}<br>
Eliminate A...<br>
{{{12A-2B=-2}}}
{{{-12A-4B=-2}}}
{{{-6B=-4}}}
{{{B=2/3}}}<br>
{{{-6A+2/3=1}}}
{{{-6A=1/3}}}
{{{A=-1/18}}}<br>
{{{-1/18+C=0}}}
{{{C=1/18}}}<br>
ANSWER:
{{{(x-2)/((x+2)^2(x-4)) = (-1/18)/(x+2)+(2/3)/(x+2)^2+(1/18)/(x-4)}}}<br>
or<br>
{{{(x-2)/((x+2)^2(x-4)) = -1/(18(x+2))+2/(3(x+2)^2)+1/(18(x-4))}}}<br>
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